Field extension which is not separable but still simple 
Let $E$ be a finite extension of a field $k$ of characteristic $p > 0$, and let $p^r = [E:k]_{i}$. Assume that there is no exponent $s <r$ such that $E^{p^s}k$ is separable over $k$ (i.e., such that $a^{p^s}$ is separable over $k$ for each $a$ in $k$). Show that $E$ can be generated by one element over $k$. (Here $[E:k]_i=\dfrac{[E:k]}{[E:k]_{s}}$, adapting Lang's notation.)

My try: Take $a\in E$. Now from the fact that there is no exponent $p^s$ with $s <r$ such that $E^{p^s}k$ is separable over $k$ we can say $[k(a):k]_i \geq p^r$. But from the fact $p^r=[E:k]_{i}$ we must have $[E:k(a)]_i=1$. So, $E$ is finite separable extension of $k(a)$. So, we can write $E=k(a,b)$. Now I'm stuck. Can anyone please help me out ?
 A: This is proof is kind of tricky, I hope someone comes up with a more convenient one:
Let $b$ be a primitive element of $E^{sep}/k$. Since $b$ is separable over $k$ it is well known that we have $k(b)=k(b^{p^n})$ for all $n > 0$. In particular $b^{p^r}$ is also a primitive element of $E^{sep}/k$.
Now take $a \in E$ such that $a^{p^{r-1}}$ is not separable $/k$ (exists by assumption). We have $a^{p^r} \in E^{sep}$ and clearly we have
$$E^{sep} = k(b^{p^r}-a^{p^r},a^{p^r})$$
By the well known proof of the "primitive element theorem", we obtain that for all but finitely many $c \in k$ the element $b^{p^r}-a^{p^r}+ca^{p^r} = b^{p^r}+(c-1)a^{p^r}$ is a primitive element of $E^{sep}/k$.
Of course we can assume $k$ to be infinite (otherweise $k$ is perfect and we have nothing to show), in particular $k^{p^r}$ is infinite. Hence we can choose $c-1$ to be a $p^r$-th power. We end up with $E^{sep} = k(b^{p^r}+d^{p^r}a^{p^r})$ with some $d \in k$.
Consider the element $x = b+da$. I claim $k(x)=E$.
$b^{p^{r-1}}$ and $d^{p^{r-1}}$ are separable over $k$, $a^{p^{r-1}}$ is not. Thus $x^{p^{r-1}}$ is not separable over $k$.
In particular we obtain for the minimal polynomial $f_x$ of $x$, that $f_x(T)=g(T^{p^r})$ for some $g \in k[T]$. We have $g(x^{p^r})=f_x(x)=0$. But $x^{p^r}$ is a primtive element of $E^{sep}/k$, hence the degree of $g$ is at least $[E:k]_s$, which finally leads us to the desired
$$\operatorname{deg}(f_x) = p^r\operatorname{deg}(g) \geq p^r [E:k]_s = [E:k].$$
A: Ok here is the shorter proof, which only uses the result of the primitive element theorem, but not its proof.
Recall the following: If $k$ is a field of characteristic $p$, we have that $x \in \overline k$ is separable over $k$ if and only if $k(x)=k(x^p)$. If $x$ is not separable, we have that $k(x)/k(x^p)$ is an extension of degree $p$.
Now to the proof:
Again, take $a \in E$, such that $a^{p^{r-1}}$ is not separable over $k$. Clearly it is not separable over $E^{sep}$ either, because otherwise we have two separable extensions $E^{sep}(a^{p^{r-1}}) \supset E^{sep} \supset k$, which would give us separability of $a^{p^{r-1}}$ over $k$.
Consider
$E^{sep}(a) \supset E^{sep}(a^p) \supset \dotsc \supset E^{sep}(a^{p^{r-1}}) \supset E^{sep}(a^{p^r})=E^{sep}$
By the above fact, all those extensions have degree $p$, in particular we get $$[E^{sep}(a):E^{sep}]=p^r=[E:E^{sep}],$$
hence $E^{sep}(a)=E$. By the primitive element theorem we have $E^{sep}=k(b)$ with $b$ separable over $k$. We obtain $E=k(a,b)$ with $b$ separable over $k$. By the primitive element theorem (stronger version!), we obtain that $E/k$ is simple.
